Visual walkthrough — RL circuit — growth and decay of current

1.8.29 · D2 · Physics › Electromagnetism › RL circuit — growth and decay of current

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Step 1 — Circuit draw karo aur har arrow ko naam do

KYA. Hamare paas ek loop hai: ek battery, ek switch, ek resistor, aur ek inductor, sab ek hi ring mein end-to-end connected hain. Bas itna hi.

WHY yahan se shuru karein. Baad ki har equation bas yahi kehti hai: "is loop ke around total voltage zero hoti hai." Yeh sentence tab tak nahi likh sakte jab tak tum loop mein har cheez jo push ya resist karti hai, use dekh nahi lete.

PICTURE. Figure dekho. Loop ko clockwise follow karo battery se start karke.

Main bata deta hoon har label ka matlab, seedhe shabdon mein:

• $\epsilon$ (battery) — ek fixed push, volts mein measure hoti hai. Ise ek pump samjho jo hamesha loop mein utni hi "electrical pressure" shove karne ki koshish karta hai. (Volt = push per unit charge.)
• $R$ (resistor) — ek friction element. Jab current $I$ isse guzarti hai, toh yeh voltage $IR$ kha jaata hai. Zyada current → zyada loss.
• $L$ (inductor) — wire ki ek coil. Yeh ziddi element hai. Iska behaviour hum Step 3 mein dhyan se banate hain.
• $I$ — current: charge kitni tez ek point se guzarti hai, amperes (amps) mein measure hoti hai. Yahi ek number hai jiska story hum bata rahe hain.

Poora page ek hi sawaal ka jawab deta hai: switch band karte hi $I=0$ se shuru hokar, $I$ time ke saath kaise badhta hai?

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Step 2 — Rate of change $\frac{dI}{dt}$, ek slope ki tarah draw kiya

KYA. Hum symbol $\frac{dI}{dt}$ introduce karte hain. Seedhe shabdon mein yeh current-versus-time graph ka slope hai — current kitne amps per second gain kar rahi hai, abhi is waqt.

WHY yahi tool chahiye, koi aur nahi. Inductor ka poora drama change ke baare mein hai: yeh react karta hai ki current kitni tez badal rahi hai, na ki woh kitni badi hai. "Ek instant par koi cheez kitni tez badal rahi hai" — is baare mein baat karne ka ek hi tool fit hota hai jo hai derivative — curve ka slope ek single point par. Yahi exactly $\frac{dI}{dt}$ hai.

PICTURE. $I$ ka $t$ ke against graph par, koi bhi instant chunno. Zoom in karo jab tak curve ek straight line jaisi na lage. Us chhoti line ki steepness hi $\frac{dI}{dt}$ hai.

Figure padhte hain:

• Jahan curve steep hai (shuruaat mein, red tangent), $\frac{dI}{dt}$ bada hai — current tezi se amps gain kar rahi hai.
• Jahan curve flat hai (baad mein, mint tangent), $\frac{dI}{dt}\approx 0$ — current muskil se badal rahi hai.

Yeh baat yaad rakho: ziddi inductor slope ko bade se shuru karke zero tak shrink karne par majboor kar dega. Woh shrinking slope hi curve ka murna hai.

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Step 3 — Inductor ka rule: yeh slope se ladta hai

KYA. Inductor apni khud ki voltage produce karta hai, jise back-EMF kehte hain, jo $L\frac{dI}{dt}$ ke barabar hoti hai. Yeh jo bhi change ho raha hai, uske khilaf point karti hai.

WHY yeh form. Faraday's law of electromagnetic induction ke anusaar, changing current ek changing magnetic field banata hai, jo ek voltage induce karta hai. Lenz's law ke anusaar, woh induced voltage hamesha us change ka virodh karti hai jisne use banaya. Size is baat ke proportional hai ki current kitni tez badal rahi hai ($\frac{dI}{dt}$) aur coil isme kitni achhi hai ($L$). Dono multiply karo: opposition $=L\frac{dI}{dt}$.

PICTURE. Do panels. Current badhne ki koshish kar rahi hai → inductor peeche dhakelta hai (ek drop). Jitna tez attempted change, utna bada backward push.

Term-by-term:

$V_L=\underset{\text{kitna ziddi}}{\underbrace{L}}\underset{\text{current kitni tez badal rahi hai}}{\underbrace{\frac{dI}{dt}}}$

• $L$ coil ki ziddigi measure karta hai. Double $L$ → same rate of change ke liye double back-push.
• $\frac{dI}{dt}$ Step 2 wali rate hai. Agar current nahi badal rahi, toh $\frac{dI}{dt}=0$, isliye inductor kuch produce nahi karta — yeh ek ordinary wire ban jaata hai.

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Step 4 — Loop ko add up karo: Kirchhoff's voltage law

KYA. Ring ke around ek baar chalo aur har voltage add karo. Total zero hona chahiye — tum jahan se shure kiye wahin wapis nahi aa sakte alag "height" par. Yahi Kirchhoff's voltage law hai.

WHY. Voltage pahadi par height jaisi hai. Kisi bhi closed path ke around jao aur tumhara net rise zero hona chahiye. Toh (battery push) minus (resistor drop) minus (inductor back-push) $=0$.

PICTURE. Loop mein battery par ek "+" gain aur do "$-$" drops, ek $R$ par, ek $L$ par.

$\underset{\text{battery uthata hai}}{\underbrace{\epsilon }}-\underset{\text{resistor khaata hai}}{\underbrace{IR}}-\underset{\text{inductor ladta hai}}{\underbrace{L\frac{dI}{dt}}}=0$

• $\epsilon$ — constant push, positive kyunki hum battery ko $-$ se $+$ cross karte hain.
• $IR$ — resistor drop; subtract kiya kyunki current isse guzarti hai aur voltage lose hoti hai.
• $L\frac{dI}{dt}$ — current badhne par inductor ki opposition ($\frac{dI}{dt}>0$), isliye yeh bhi ek drop ki tarah kaam karta hai.

Ab akele slope ke liye rearrange karo:

$\frac{dI}{dt}=\frac{\epsilon -IR}{L}$

Ise slope ke liye ek rule ki tarah padho: kisi bhi current $I$ par, curve ka slope $(\epsilon -IR)/L$ ke barabar hai. Yeh akeli line shape ki poori kahani pehle se bata deti hai — Step 5 ise padh leta hai.

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Step 5 — Slope rule se seedha shape padho

KYA. Kisi bhi integration se pehle, bas $\frac{dI}{dt}=\frac{\epsilon -IR}{L}$ ke baare mein do siron par sochte hain.

WHY. Ek achha physicist pehle equation se qualitative shape padhta hai, phir algebra se confirm karta hai. Yahi se bending aati hai.

PICTURE. Curve jisme shuruaat, beech aur ant mein slope arrows draw kiye hain.

• Shuruaat mein ($t=0$, toh $I=0$): $\frac{dI}{dt}=\frac{\epsilon -0}{L}=\frac{\epsilon }{L}$ — sabse bada slope. Curve origin se steeply nikalti hai.
• Jaise $I$ badhta hai, term $IR$ bada hota jaata hai, toh $\epsilon -IR$ shrink hota hai, toh slope shrink hota hai — curve murr jaati hai.
• Ant mein, jab $IR=\epsilon$, top aur slope dono zero ho jaate hain: $\frac{dI}{dt}=0$. Current badhna band ho jaata hai. Yeh special value hai

$I_{\max }=\frac{\epsilon }{R}.$

Toh bina solve kiye bhi hum jaante hain: steep se shuru, murr ke, $\epsilon /R$ par flat. Yeh ek exponential approach hai.

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Step 6 — Slope rule ko curve mein badlo (integration)

KYA. Ab hum $\frac{dI}{dt}=\frac{\epsilon -IR}{L}$ solve karte hain taaki $I$ ko $t$ mein ek formula ke roop mein paa sakein.

WHY integration. Derivative ne hume har point par slope diya. Actual curve ko uske slopes se rebuild karne ke liye, hum derivative ko ulta chalate hain — woh operation integration kehlaata hai (saare chhote slope-steps ko jodhna). Yeh naturally $\frac{d}{dt}$ ka undo hai.

Kaise — variables alag karo. $I$ wali sab cheez left par rakho, $t$ wali sab cheez right par, taaki har side apne aap add ho sake:

$\frac{dI}{\epsilon -IR}=\frac{dt}{L}$

Switch-on se ($I=0$ at $t=0$) ek general baad ke moment tak ($I$ at $t$) add up (integrate) karo:

${\int }_0^I\frac{dI}{\epsilon -IR}={\int }_0^t\frac{dt}{L}$

Right side easy hai: $\frac{t}{L}$. Left side, jo parent note mein work kiya gaya hai, deta hai $-\frac{1}{R}\ln \left(\frac{\epsilon -IR}{\epsilon }\right)$. Dono barabar rakh ke:

$\ln \left(\frac{\epsilon -IR}{\epsilon }\right)=-\frac{Rt}{L}$

• $\ln$ natural logarithm hai — yeh jawab deta hai "$e$ ko kaunsi power pe raise karein toh yeh mile?" Hum ise isliye use karte hain kyunki $1/(\text{stuff})$ integrate karne par hamesha log aata hai.
• Right side mein combination $\frac{R}{L}$ hai, jiska unit $1/\text{seconds}$ hai — iska reciprocal $\tau =L/R$ ek time hai. Yahi timescale hai jo hum aage naam denge.

PICTURE. Left integral $\frac{1}{\epsilon -IR}$ curve ke neeche shaded area ki tarah dikhaya gaya hai — "slopes ko add karna."

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Step 7 — Log undo karo: exponential appear hota hai

KYA. Logarithm hatao dono sides ko $e$ ki power par uthaakar. Exponential $e^x$ exactly woh function hai jo $\ln$ ko undo karta hai.

WHY exponential. $\ln$ aur $e^{(\cdot )}$ perfect opposites hain: $e^{\ln x}=x$. $e$ apply karne se $I$ log ke andar se nikal aata hai.

$\frac{\epsilon -IR}{\epsilon }=e^{-Rt/L}$

$I$ ke liye solve karo aur $\tau =L/R$ likho:

$\boxed{I(t)=\frac{\epsilon }{R}(1-e^{-t/\tau })},\tau =\frac{L}{R}$

Term-by-term, jahan har cheez rehti hai:

$I(t)=\underset{\text{ceiling }{I}_{\max }}{\underbrace{\frac{\epsilon }{R}}}(1-\underset{\begin{array}{c}\text{1 se shuru,}\\ \text{0 par khatam}\end{array}}{\underbrace{e^{-t/\tau }}})$

• $\frac{\epsilon }{R}$ — Step 5 wali flat ceiling.
• $e^{-t/\tau }$ — ek factor jo $t=0$ par $1$ hai aur $0$ ki taraf decay karta hai. Toh $(1-e^{-t/\tau })$ $0$ se $1$ tak chadha rehta hai.
• $\tau =L/R$ — time constant: poore process ki clock speed.

PICTURE. Exponential term $e^{-t/\tau }$ girta hua, aur uska mirror $1-e^{-t/\tau }$ chadha, dono saath plot kiye gaye hain jahan $t$ ko $\tau$ ke units mein measure kiya gaya hai.

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Step 8 — $\tau =L/R$ pace kyun set karta hai (edge behaviours)

KYA. Do extreme circuits dekho — bahut ziddi ($L$ huge) aur bahut lossy ($R$ huge) — aur degenerate case $R\to 0$ bhi.

WHY. Ek formula jisme tum sirf plug in kar sako, woh aadha samjha hua hai. $L$ aur $R$ ko extremes tak push karne se pata chalta hai ki $\tau =L/R$ kyun hai, $LR$ ya $R/L$ kyun nahi.

PICTURE. Ek axis par teen growth curves: bada $\tau$ (slow, lavender), medium (coral), chhota $\tau$ (fast, mint).

• Bada $L$ → bada $\tau$ → coil bahut ziddi hai → current dheere chadha rehta hai (lavender curve right mein stretch).
• Bada $R$ → chhota $\tau$ → strong friction cheezein jaldi settle karti hai (mint curve jaldi upar snap ho jaati hai).
• Units check (parent ka mistake-buster): $[L]/[R]=\text{H}/\Omega =(\text{V⋅s/A})/(\text{V/A})=\text{s}$ ✓. $LR$ galat units deta, toh $\tau =L/R$ forced hai.
• Degenerate case $R\to 0$: tab $\tau =L/R\to \infty$ aur $I_{\max }=\epsilon /R\to \infty$. Formula kehta hai current slope $\epsilon /L$ ke saath hamesha ke liye badhti rehti hai (ek straight line, koi ceiling nahi). Physically: koi resistor nahi jo use cap kare, toh ideal battery current ko bina bound ke upar drive karta hai — curve kabhi flat nahi hoti.

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Ek-picture summary

Upar sab kuch compressed: left mein slope rule (steep→flat), beech mein integration, right mein finished exponential curve, teen landmark dots (0, 63.2%, ~100%) marked.

Recall Poore walkthrough ki Feynman retelling

Switch band karo. Battery current push karna chahti hai, lekin coil achanak change se nafrat karti hai, toh pehle instant mein current abhi bhi zero hai — curve floor se shuru hoti hai. Battery ki poori push kahin aur jaane ke liye nahi hai siwaaye coil mein, toh current apne sabse steep angle par chadha rehta hai bilkul shuruaat mein. Jaise current banta hai, resistor battery ki push ka zyada se zyada hissa khaane lagta hai ($IR$ badhta hai), aur speed badhane ke liye kam push bachti hai. Toh chadhai aur aur aalsani hoti jaati hai. Aakhirkar resistor poori battery push kha jaata hai; ab current badlane ke liye kuch nahi bacha, coil chup ho jaati hai, aur current $\epsilon /R$ par flat reh jaata hai. Mathematically woh "steep se shuru, murr ke, flat" ek exponential approach hai, aur ek number — $\tau =L/R$ — stopwatch hai: ek $\tau$ ke baad tum 63.2% upar ho, lagbhag paanch $\tau$ ke baad basically wahan pahunch jaate ho. Badi coil, dheeri chadhai; bada resistor, tezi se khatam. (Capacitor ki mirror-image kahani ke liye dekho RC circuit — charging and discharging; coil ki energy kahan gayi, ke liye dekho Energy stored in a magnetic field.)

Recall Pictures par quick self-check

• Step 5 mein, slope kyun shrink hota hai jaise current badhti hai? ::: Kyunki slope $=(\epsilon -IR)/L$ hai; jaise $I$ badhta hai, $IR$ badhta hai, toh $\epsilon -IR$ shrink hota hai.
• Step 6 mein curve ko uske slope se rebuild karne wala operation kaunsa tha? ::: Integration (derivative ka undo).
• Step 7 mein dono sides par $e$ raise karna kyun aata hai? ::: $e^{(\cdot )}$ $\ln$ ko undo karta hai, $I$ ko logarithm ke andar se azaad karta hai.
• Agar $R\to 0$ ho toh curve ka kya hota hai? ::: $\tau \to \infty$, koi ceiling nahi — current slope $\epsilon /L$ ke saath hamesha ke liye straight line ki tarah badhti rehti hai.

Connections

• Kirchhoff's voltage law — Step 4 mein loop sum.
• Inductance and self-induction — $LdI/dt$ term ka source (Step 3).
• Faraday's law of electromagnetic induction — changing current voltage kyun induce karta hai.
• Lenz's law — woh induced voltage change ka virodh kyun karta hai.
• Energy stored in a magnetic field — built-up current ki energy kahan baithe hai.
• RC circuit — charging and discharging — same-shaped curve, capacitor mirror role mein.

Concept Map